Partial Fraction Decomposition Calculator

It's the process of rewriting one rational expression, a polynomial divided by another polynomial, as a sum of simpler fractions that add back up to the original. Each simpler fraction has a denominator that's just one factor (or one power of a factor) of the original denominator, which makes the result far easier to integrate, transform, or otherwise work with than the combined fraction was.

Factor the denominator completely into linear and irreducible quadratic pieces, assign the matching unknown form to each factor (a constant for a distinct linear factor, a linear numerator for a quadratic one, extra terms for any repeated factor), multiply both sides by the full denominator, and compare coefficients of each power of x to build a solvable system of equations for every unknown.

Most often when integrating a rational function in calculus, since a decomposed sum of simple fractions integrates directly into logarithms and arctangents while the combined fraction usually doesn't have an obvious antiderivative. It's equally common in engineering when taking an inverse Laplace transform, since transform tables are built from these same simple forms rather than complicated combined fractions.

The distinct-linear-factor case is genuinely quick once the cover-up method clicks. It gets noticeably harder with a repeated factor or an irreducible quadratic, mainly because it's easy to under-count how many unknowns the setup actually needs. Most of the real difficulty is in the setup, correctly identifying every factor's type and multiplicity, rather than in solving the system once it's written down correctly.

Distinct linear factors, repeated linear factors, distinct irreducible quadratic factors, and repeated irreducible quadratic factors. Every denominator that comes up in practice is built from some combination of these four, and the unknowns needed for the decomposition depend entirely on which types are present and how many times each factor repeats.

A shortcut for distinct linear factors: after multiplying both sides by the denominator, plug in the specific x-value that makes one factor zero, which cancels out every other unknown and leaves a single one-step equation for that unknown alone. Repeating this for each distinct linear root solves for all of them individually, without ever building or solving the full coefficient system.

Yes. A linear factor raised to the power k gets one term for every power from 1 up to k, and a repeated irreducible quadratic follows the same rule with a linear numerator at each step. Leaving out any of those terms is the single most common setup mistake, since it leaves the system with fewer unknowns than the denominator's degree actually requires.

When the numerator's degree is equal to or greater than the denominator's, the fraction is improper and needs polynomial long division first. That division produces a whole polynomial quotient plus a proper remainder; only the remainder gets decomposed into partial fractions, and the quotient is added back onto the final answer.

It decomposes over the real numbers, the form used in essentially all calculus and engineering coursework, where an irreducible quadratic factor is kept as one quadratic term rather than split into a complex-conjugate pair. It does not output complex-number coefficients directly, since real textbook and homework problems almost never ask for that form.

Yes, a denominator with one or two distinct linear factors, the simplest and most common homework case, decomposes the same way as any harder one, just with fewer steps and fewer unknowns to solve for. The calculator shows the same full step-by-step breakdown regardless of how simple or complicated the denominator turns out to be.

Yes, the reference table above lists the exact decomposition form for each of the four factor types side by side with a real example of each. It's the same rule set explained in the guide sections on this page, just condensed into a single lookup table rather than a downloadable file.

Because that's a different question: it's asking how to solve the resulting system of equations on a specific physical calculator model during an exam, not what the decomposition itself is. This tool solves the actual math directly and shows every step, which works regardless of what calculator, if any, someone has in hand.

Up to degree 12, with whole-number coefficients in both the numerator and denominator. That comfortably covers real coursework and engineering problems, which rarely go past degree 6, while leaving real margin above that before floating-point precision genuinely becomes a limit.

Yes. Typing (x-2)(x+1) works exactly the same as typing its expanded form, x²-x-2, and a leading number like 2(x-1)(x+2) works too, including a repeated factor written as (x+1)^2. This calculator multiplies everything out internally before factoring, so entering a denominator either way gives the identical decomposition and the same step-by-step answer.